A wire of resistance 5 ohm is drawn out so that its new length is three times its original length. What is the resistance of the new wire?

5/3 ohm

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5 ohm

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15 ohm

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45 ohm

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Solution

The correct option is D 45 ohm
Let m be the mass of the given wire, d its density and p its resistivity. These quantities remain unchanged when the wire is drawn out. Let $l_{1}$ be the length of the smaller wire and $r_{1}$ its radius of cross section and If $l_{2}$ and $r_{2}$ represent the length and radius of cross section of the elongated wire we have
$m = π r_{1}^{2} l_{2} d = π r_{2}^{2} l_{2} d$
$r_{1}^{2} l_{1} = r_{2}^{2} l_{2}$
It is given that $l_{2} = 3 l_{1}$
Therefore $r_{2}^{2} = \frac{r_{1}^{2}}{3}$
If $R_{1}$ is the resistance of the smaller wire and $R_{2}$ that of the elongated wire we have $R_{1} = \frac{p l_{1}}{π r_{1}^{2}}$

$R_{2} = \frac{p l_{2}}{π r_{2}^{2}}$

$\frac{R_{2}}{R_{1}} = \frac{l_{2}}{l_{1}} \times \frac{r_{1}^{2}}{r_{2}^{2}} = 9$
$R_{1} = 5 o h m$
$R_{2} = 45 o h m$

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